* 807 
py 1 



NOTES ON MECHANICS 



PREPARED FOR 



THE USE OF THE STUDENTS 



-OF- 



The Pennsylvania State College 






-by- 



ELTON D. WALKER, 

Professor of Hydraulic and Sanitary Engineering. 



NOTES ON MECHANICS 



Prepared for the use of the Students 



-of- 



THE PENNSYLVANIA STATE COLLEGE 



-by- 



ELTON D. WALKER, 
Professor of Hydraulic and Sanitary Engineering. 



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STATE COLLEGE, PA. 

NITTANY PRINTING AND PUBLISHING CO. 

1903. 



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THE. LIBRARY OF 
CONCRESb. 

T wo Copies Receives 

OCT 12 1903 

Copyright Entry 

CLASS Ql/ XXc No | 

"1 X c> 3 
COPY 



Copyright, ir03, by Elton I). Walker. 



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NOTES . ON MECHANICS. 



Parallel Forces in Space. — Refer the forces to a set of 
rectangular co-ordinate axes such that the Z axis is parallel to the 
forces. 

Denote the forces by P lt P 2 , P 3 , etc., and their co-ordinates 
by x l y v x 2 y 2 , etc., and the resultant by R and its co-ordinates 
by XY. 

Then, the magnitude of the resultant is equal to the algebraic 
sum of the forces, or R=2P. 

The moment of the resultant about OY is equal to the sum of 
the moments of the forces about OY, or X2P=2Px. 

For a similar reason, Y2P=SPy. 

u Y 2Px SPy 
Hence, X= and Y= 

2P 2P 

It may also be proven, that in the case where the forces are 
not parallel to either of the rectangular axes, 

R=2P , x=^ x , Y =*L y . z=*5. 

2P 2P 2P 

X, Y, and Z being the co-ordinates of a point in the line of action 
of R. 

Note. — In this case the products Px, Py, Pz, etc., are not 
moments. 

Centre of Gravity.— One particular case of parallel forces 
is found in the attraction exerted by the force of gravity upon the 
different particles of any body. For convenience, we speak of the 



4 NOTES OX MECHANICS. 

center of gravity of surfaces and of lines, although such surfaces 
and lines have no mass, meaning by that term the point of appli- 
cation of the resultant force which would be in equilibrium with a 
system of parallel forces uniformly distributed over such surface 
or along such line. This leads us to the following definition. The 
center of gravity of a body, surface, or line is the point of applica- 
tion of a resultant force which will hold in equilibrium a system of 
parallel and uniformly distributed forces acting upon the body, 
surface, or line. 

If X, Y, Z denote the co-ordinates of the center of gravity of 
any body referred to three co-ordinate axes and if we denote the 
volume of an element by dV and its heaviness or weight per unit 
volume by w, and the co-ordinates of the element by x, y, z; then 
using the integral sign to indicate the summation of like terms for 
all particles of the body, we have, for heterogeneous bodies, 



/wxdV I wydV j \ 

C wdV f wdV f 



If the body is homogeneous, w is constant and cancels out 
leaving 

fxdV fydV ( zdV 

X=^ , Y=~ , Z ' 

V V V 

We may consider a surface to be an infinitely thin homoge- 
neous shell of uniform thickness and if dA denote an element and 
A the whole area of the surface, the above equations become 



xdA I ydA ( zdA 

, Y % - , Z ' 

AAA 

Similarly for a homogeneous wire of constant, small cross-sec- 



NOTES ON MECHANICS. 5 

tion, and for its limit, a geometric line, its length being s and an 
element of length, ds, we obtain 



J xds I yds I 2 



zds 

Y=* , Z=*' 

s s s 

Problem 1. — Required the position of the center of gravity 
of a circular arc AB. Fig. 1. Take the origin O at. the center of 
the circle, and the X axis bisecting the arc. Let the length of 
the arc be s and let ds denote an element of the arc. We need 
determine only X, since Y=0 from the conditions of symmetry. 

I xds 

X=- . 

s 

From similar triangles we have 

ds : dy : : "r : x, 

rdy r /»+ a 2ra . 

therefore ds— — , and X— - 8 dy =• — » L e -» 
x s J _ a s 

equals the chord X the radius -+- the length of the arc. For a 

semicircular arc this reduces to X=2r-^7r. 

Problem 2.— Center of gravity of trapezoids and triangles. 

Fig. 2. Prolong the non-parallel sides of the trapezoid to intersect 

at O, which take as the origin, making the X axis perpendicular to 

the bases b and b r Taking a vertical strip as our element of area, 

b b 

we have for its height, -,- x, and for its area, dA=-j— xdx. 



Now A=-~(bh — bjhj ^—(hMi, 2 ) and X= becomes 



b i xdA 



b 



J x 2 dx 



x= h J M 2 V-b? 



lb 3 h 2 -h* 

(h 2 -h 2 ) l 

2 h v ' ; 



for the trapezoid. 



6 NOTES ON MECHANICS. 

2 
For a triangle h^O, and we have X=-«-h ; that is, the 

center of gravity of a triangle is one third the altitude from the 
base. The center of gravity is finally determined by knowing that 
a line joining the middles of the parallel sides of the trapezoid, or 
the vertex to the center of the base of a triangle, passes through 
the center of gravity. 

Problem 3. — Sector of a circle. Fig. 3. Let the angle of the 
sector =2a. Using polar co-ordinates, the element of area 
dA=pdp.d<f>, and its x=p cos <f> ; hence the total area 



/■ 

ftlf r o ™s<i>p 2 dpd<f> 

tJ -a C ° S * ^ 



r a. 



X=-£- I xdA 

1 

1 



~A3 
2 r sin a 

4 r sin }4/3 
or, putting /3=2a, x =^ ~ 

4r 
For a semicircle this reduces to X=-^— . 

Problem 4. — Sector of a flat ring, or annulus. The treat- 
ment is similar to that of the last case, the difference being that 

r. r 



the limits of integration are 



instead of 
r_ 



, giving as the result 



o 



4 r/-r t » sin HP 
3 r^-r, 1 fi 

Problem 5. — Segment of a circle. Fig. 4. Since each rec- 
tangular element of any vertical strip has the same x, we may 



NOTES ON MECHANICS. 7 

take these vertical strips as the elementary areas. Then 
dA=2ydx. From similar triangles x : y : : dy : dx, i. e., xdx=ydy. 

fxdA fx2ydx 2fydy , 

Hence, X== ~_ rL ^_« _ JL 

A A A ~~ 3 A 

but a = the half chord, hence, X = i — 2_. 

12A 

Problem 6. — Homogeneous cone or pyramid. Divide the 
body into laminae by means of planes parallel to the base and ap- 
ply the principles of the preceding cases. 

Centrobaric Method of Determining Areas and Vol= 
umes. — If an elementary area, dA, be revolved about an axis in 
its plane, but not included in its area, through the angle, a, not 
greater than 2tt, the distance of the elementary area from the 
axis of revolution being x, the volume generated is dV=axdA, 
and the total volume generated by all the elementary areas of a 
finite plane figure which lies entirely on one side of the axis of 
revolution and whose plane contains that axis is 

V= fdV= a fxdA; 

but a J xdA =aAX, and hence 

aX being the length of the path described by the center of gravity 
of the plane figure, we may write: The volume of a solid of revo- 
lution generated by a plane figure, lying on one side of the axis, 
equals the area of the figure multiplied by the length of the curve 
described by the center of gravity of the plane figure. 

A similar proposition may be deduced for the surface generated 
by the revolution of a line. 

The angle, a, must be expressed in radians for numerical work. 

Non=concurrent Forces in Space. — Fig 5. Let P lf P 2 , 

etc., be the given forces, and x 1 y 1 z 1 , etc., the co-ordinates of 
their points of application referred to an arbitrary origin and axes; 



8 NOTES ON MECHANICS. 

a, /3, y v etc., the angles made by their lines of action with X, Y, 
and Z. 

Considering the first force, P lf replace it by its three compo- 
nents parallel to the three axes. X 1 =P ] cos a,, Y 1 =P 1 cos ff v 
Zj^Pj cos y v Pj itself is not shown in the figure. At and 
also at A put a pair of equal and opposite forces, each equal and 
parallel to Z v Z 5 at B is now replaced by the single force Z x 
acting upward at the origin, and two couples, one in a plane paral- 
lel to YZ and having a moment equal to — ZjV,, and the other in a 
plane parallel to XZ and having a moment equal to -f-ZjXj. Simi- 
larly at O and C put in pairs of forces equal and parallel to X, and 
we have X } at B replaced by the force X 1 at the origin and two 
couples, one in a plane parallel to XY and having a moment 
-j-XjV,, the other in a plane parallel to XZ and having a moment equal 
to — XjZjj and finally by a similar device, Y, at B is replaced by 
the force Y, at the origin and two couples parallel to XY and YZ 
and having moments, — Y,x, and -f-^z,, respectively. 

We have therefore replaced the force P, by three forces X 1( 

Y ]f and Z x at and six couples. Combining each pair of couples 

whose axes are parallel, they can be reduced to three, viz.; 

One with X axis and a moment equal to YjZ, — Z,y,; 

One with Y axis and a moment equal to Z,Xj — X,^; 

One with Z axis and a moment equal to Xj, — Y,x,. 

Dealing with each of the other forces, P.,, P , etc., in the same 
manner, the whole system may be finally replaced by three 
forces, 2X, 2Y, 2Z, at the origin and three couples whose mo- 
ments are respectively, 

L--2(Yz — Zy) with its axis parallel to X; 
M 2(Zx — Xz) with its axis parallel to Y; 
N = 2(Xy — Yx) with its axis parallel to Z. 

The axes of these couples being parallel to the respective co- 
ordinate axes, X, Y, and Z, and proportional to the moments, L, 
M, and N respectively, the axis of their resultant whose moment is 



NOTES ON MECHANICS. 9 

G, must be the diagonal of a parallelopiped constructed on the three 
component axes proportional to L, M, and N. Therefore 

G= V L'+M'-fN" and the resultant of 2X, Xf, and 2Z is 
R= l/(2X) 2 +(2Y) 2 +(2Z) 2 acting through the origin. 
If a, (}, y are the direction angles of R, we have 
2X 2Y 2Z 

COS a= -b» COS /3 = _j-, COS y —~^/> 

and if A., fx, v, are the direction angles of the resultant couple G, 
we have 

, L M N 

COS \=-—, COS fi= -, COS Vz=— . 

G G G 

For equilibrium, G=0 and R=0, or stated separately, 

2X=0, Tf=0, 2Z=0, L=0, M=0, and N=0. Hence, if a 

system of non-concurrent forces in space is in equilibrium, the 

plane systems formed by projecting the given system upon each 

of three arbitrary co-ordinate planes will each be in equilibrium. 

Statics of Flexible Cords. — In what follows, all cords are 
supposed to be perfectly flexible and inextensible and all forces are 
supposed to be in the same plane. All of the deductions are 
based upon the following axioms : 

First, the strain on any cord at any point can act only along 
the cord, or along the tangent if the cord be curved. 

Second, we may apply to flexible cords in equilibrium all the 
conditions for equilibrium of rigid bodies. 

Third, conditions of equilibrium can not be applied unless the 
system can be considered a free body. This is allowable only 
when we conceive the supports to be removed and the reactions 
exerted by the supports or fastenings put in. 

These reactions having been put in, let us consider the case 
shown in Fig. 6. If we take any point, A, on the cord as a center 
of moments, knowing that the resultant, R, of the forces, P,, P 2 , P g , 
situated on one side of A must act along the cord through A, we 
have 



10 NOTES ON MECHANICS. 

o ,RYO-0 That is, in a system of 
PA— PA— P A ^^ t of mom ents be taken 

situated on either side of this point will equal zero. 

on A cord in equilibrium over a pulley whose axle s 

Pulley.— A cora M . . . Fi „ 7. For, consid- 

smooth, has the same tenston on hot! side* F g ^ 

erin g the puUey and its J*£- «*£ J£ cQrd . Hence 

T" "lute R^ *e axle bisects the angle, a, and therefore, if a 

the pressure, K, at me aAi position of 

weighted pulley rides ^/J^^/^L through A by 

^ m Ti:c»^°^ «* *• - d and center at 

C a! men ££ n^ont a, through the middle of AO cutting 

CDinB ' t_ c- *A Knots— Given all the 

We ;^emXf d ^ 9 b Ind F, on: d wefght°o;, required the re 
geometric elements «r*' the oint3 f 

maining weights and the forces H, V H .. ^ 

support that equilibrium may exist H ; and ^ and slmI . 

and vertical components o, the tension m f,^ ^ 
larly H D and V n , those at n. There are 
the second axiom we have 

SY==0j that is, H o -H n ^O and 

rGfG-f )— (v+v a )=a 

While from the third axiom, taking 'the successive knots, 1 , 2. 

etc., as centers of moments, we have 

-VA+Hjfr^. 

-Va+H^+G^x,— x l )=0, 

-V x 3 +H o y 8 +G 1 (x a — x 1 )4-G > (x,-x a )-u ) 

etc for the n knots. Thus we have n+ 2 independent equation, 

and can solve for the unknown quantities. 

r, u~\ a If the weights are equal and 
Loaded Cord as a Parabola.-K t lie g 

infinitely small and are uniformly spaced along the 



NOTES ON MECHANICS. 11 

equilibrium exists, the cord, having no weight, will form a para- 
bola. In Fig. 10, let w equal the weight of the loads per linear 
unit ; let be the vertex of the curve and M any point on the 
curve. We may consider the portion OM as a free body if the 
reactions of the contiguous portions of the cord are put in, H o and 
T, and these according to the first axiom must act along the tan- 
gent to the curve at and M respectively. That is, H o is nod- 
dy 
zontal and T makes some angle, <f>, whose tangent equals -p- , with 

the axis X. Applying the second axiom, 2X=0 gives 

dx 
T cos <f> - H o = 0, i. e., T-^^H,. ( 1 ) 

dy 
2Y=0 gives T sin <f> — wx=0, i. e., T-j— =wx. (2) 

dy wx 
Dividing (2) by ( 1 ), member by member, we have -j—=-tt. 



dx 
Therefore dy=wxrr- is the differential equation of the curve. 



w r x 


2H 




x 2 = 2H — 

o W 





or, 

which is the equation of a parabola whose vertex is at and whose 

axis is vertical. 

dy wx 
Note. — The equation,-,— = ~rr, ma y a ^ s0 De obtained by consid- 
er 
ering that we have a free body, Fig. 1 1 , acted on by three forces, 

T, H o , and R = wx acting vertically through the middle of the 

abscissa, x. The resultant of H o and R must be equal and oppo- 

R dy wx 
site to T. Therefore tan <£=tt or ^-=tf • Evidently also, 

O 

the tangent line bisects the abscissa, x, which is a property of the 
parabola. 



12 NOTES ON MECHANICS. 

The Catenary. — A flexible, inextensible cord or chain of 
uniform weight per unit of length hung at two points and supporting 
its weight alone forms a curve called a catenary. Let the tension, 
H o , at the lowest point or vertex be represented, for convenience, 
by the weight of an imaginary length, c, of similar cord weighing 
w pounds per unit of length, i. e., H o =wc. A portion of cord of 
length, s, weighs ws pounds. Fig. 12 shows as free and in equi- 
librium a portion of the cord of any length, s, measured from the 
vertex. The load is placed uniformly along the curve and not as 
in the last section. 

dy dx 

2Y— gives T j- = ws. 2X=0 gives T,- = wc. Hence, by 

division, cdy=sdx and squaring, c 2 dy 2 =s 2 dx 2 . ( 1 ) 
Put dy 2 =ds 2 — dx 2 and we have, after solving for dx, 

cds A ds 

dx=— -=— 2 ; therefore x=rc I --,— -„ 

V S /J T& J o FS 2 +C 2 



. s+vV+c 2 
and x = c log (2) 

which shows the relation between the horizontal projection and the 
length of the curve. 

Again in ( 1 ) put dx 2 =ds 2 — dy 2 and solve for dy. 

sds 1 d(c 2 +s 2 ) 

This gives dy ^ v - T7=Y(cH ^ ya . 

Therefore y=4r P(c 2 f s 2 )~^ d(c 2 +s 2 ) 
2 •/ 

and finally y^i/j^fc 2 — c. (3) 

Solving for c, .we have c=(s 2 — y 2 ) ;-2y. (4) 

Direct Central Impact. — Suppose two masses, m, and m.„ 
to be moving in the same straight line so that the distance be- 
tween them is constantly diminishing, and that when collision or 
impact occurs, the line of pressure between the two bodies co- 
incides with the line connecting their centers of mass. Such a 



NOTES ON MECHANICS. 13 

collision is called direct central impact and the motion of each 
mass during contact is variably accelerated and rectilinear, the 
only force acting upon it being the pressure of the other body. 
While the bodies are in contact, the pressure between them grad- 
ually increases and the bodies are compressed, their centers of 
mass approaching each other until the pressure and consequent 
compression attain a maximum, at which instant the two bodies 
move with a common velocity. After this, if the bodies possess 
any elasticity, the pressure continues, but gradually reduces to zero 
when the contact ceases and the bodies separate with different 
velocities. 

Reckoning time from the first instant of contact, let t' equal the 
duration of the first period of contact, and t" that of the first plus 
the second. Let m l and m 2 , Fig. 13, be the masses and v x and 
v 2 , simultaneous velocities of the two centers of mass at any in- 
stant during contact. Let P be the variable pressure between the 
two bodies. At any instant the acceleration of m x is i x = — P-j-m lf 
and that of m 2 is f 2 = + P-^m 2 , m x being retarded and m 2 ac- 

dv 
celerated in velocity. In general, f = — . 

dt 
Therefore, m 1 dv 1 = — Pdt and m 2 dv 2 = + Pdt. (1) 

Summing all similar terms for the first part of the impact, let- 
ting the velocities before impact be u x and u 2 and the common 
velocity at the instant of maximum pressure, w, we have 
w t' t' 

m l Cdw 1 = - f Pdt, or m^w-iij) = — f Pdt. (2) 

J Ui J J 

•»w M V 

m 2 fdv 2 = -f (Pdt, or m 2 (w— u 2 ) = -f- fpdt. (3) 

J u 2 J O J 

Eliminating the second members of (2) and (3) and solving 

m ii+ m u 
for w, we obtain w— , (4) 



14 NOTES ON MECHANICS. 

which is the common velocity at the instant of maximum pressure. 

If the impact is inelastic, the bodies do not separate but con- 
tinue to move with a common velocity. 

Suppose that the impact is partially elastic; that the bodies are of 

ri" 

the same material ; and that the summation I , Pdt for the sec- 



ond period of impact bears a ratio, e, to that of | Pdt already 



Si 

J o 



used. If the impact is not too severe, we have, summing equa- 
tion ( 1 ) for the second period and letting v/ and v 2 ' equal the veloc- 
ities after impact, 

y' t" t' 

m, f dv, = — f Pdt, i. e., m,(v/— w)=— e f Pdt ; (5) 
J w J M J o 

v' t" t' 

and m 2 f dv 2 =+ f Pdt, i. e., m 2 (v 2 '— w) =+e f Pdt. (6) 
J w J t' J o 



Having determined the value of j Pdt from (2) and (3) ii 



e is called the coefficient of restitution. 

.t' 

>dt from (2) and (3) i 
o 

terms of the mass and initial velocities, substitute it and that of w 
from (4) in (5) and (6) and we have for the final velocities, 

, _ "VVf m 2 u— em 2 ( u— u 2 ) 

v '~ m 1 +m 2 V> 

m 1 u 1 -f-m,u.,+em.(u. — u,) 
and v/=- 1J ^ , lV ^ -■ (8) 

If e=0, i. e., for inelastic impact, v,'— v 2 '=w in equation (4). 
If e— 1, i. e., for elastic impact, (7) and (8) become somewhat 
simplified. 

To determine e experimentally, let a ball of the substance fall 
upon a very large slab of the same substance, noting both the 
height of the fall, h, and the height of the rebound, H ; then re- 



NOTES ON MECHANICS. 15 

garding the mass of the slab as infinite compared with the mass 

of the ball, we obtain e = A /_ 

\h' 

Virtual Velocities. — If a. material point is moving in any 
direction not coincident with that of the resultant force acting, and 
any element of its path, ds, be projected upon the line of action of 
the force, the length of this projection, du, is called the virtual ve- 
locity of the force. The product of a force by its virtual velocity 
is called its virtual moment or virtual work and is reckoned posi- 
tive or negative according as the direction of the virtual velocity 
is the same as that of the force or not. 

Proposition 1 . — The virtual work of a force equals the algebraic 
sum of the virtual works of its components. Fig. 1 4. Take the direc- 
tion of ds as the axis X ; let P l and P 2 be the components of P ; 
a,, a 2 , a, their angles with X. Then P cos a— P, cos c^-f P 2 cos a 2 . 

Hence, P ds cos a=P, sd cosa 1 -fP 2 ds cosa 2 ; but ds cos a 
equals the projection of ds upon P, i. e., equals its virtual velocity, 
du. Therefore, P du=P t du,-f-P 2 du,. If in Fig. 14, a, were 
greater than 90°, evidently we would have P du= — P, dUj-f-Pjdu^ 
i. e., Pj du t would then be negative and OD, would fall behind 
0. Hence the definition of positive and negative as above given. 

This proof is equally applicable whether all the forces are in the 
same plane or not. 

Proposition 2. — The sum of the virtual works equals zero for 
concurrent forces in equilibrium. The resultant force is zero, 
hence from the preceding proposition, 2(Pdu)— 0. 

Proposition 3. — The sum of the virtual works equals zero for 
any small displacement or motion of a rigid body in equilibrium 
under non-concurrent forces in a plane, all points of the body 
moving parallel to this plane. 

1st. Let the motion be a translation, all points of the body de- 
scribing equal and parallel lengths equal to ds. Fig. 15. Take 



16 NOTES 6N MECHANICS. 

X parallel to ds; let a,, etc., be the angles of the forces with X. 
Then 2(P cos a) = 0. Therefore ds 2(P cos a) = 0, but 

ds cos a,=du,; ds cos o 2 =du 2 ; etc. Therefore 2(Pdu)=0. 

2nd. Let the motion be a rotation through a small angle, dO, in 
the plane of the forces about any point, 0, in that plane. Fig. 16. 
With O as a pole let p x be the radius vector of the point of appli- 
cation of P, and ^ its lever arm from O. Similarly for the other 
forces. In the rotation, each point of application describes a small 
arc, ds,, ds 2 etc., proportional to p x , p 2 , etc., since ds, = p,d0, 
ds 2 =p 2 d$, etc. 

P, a,+ P 2 a 2 + ... .=0, 

but from similar triangles, ds, : du, : : p x : a,, 

p.du du. 

therefore a. =H — — t^- 
1 ds, d$ 

du. 
Similarly a 2 == ^ • etc - 

Hence, we must have P i du *+ P 2 du *+ : • • = p t i. e ., 2(Pdu)=0. 

Now since any small displacement or motion of a body may 
be conceived to be accomplished by a small translation followed 
by a rotation through a small angle, and since the foregoing deals 
only with the projection of paths, the proposition is established and 
is called the principle of virtual velocities. 

Generality of the Principle of Virtual Velocities. — 

If any mechanism of flexible, inextensible cords, or of rigid bodies 
joined together, or of both, at rest, or in motion with very small 
accelerations, be considered collectively, or any portion of it, and 
all external forces be put in, then, disregarding friction, for a 
small portion of its prescribed motion 2(Pdu) must equal zero. 

When the acceleration of the parts of the mechanism is not 
practically zero, 2(Pdu) will not equal zero, but some function of 
the mass and velocities. 



NOTES ON MECHANICS. 17 

MOMENT OF INERTIA. 

Rigid Bodies. — The moment of inertia of a rigid body about 
any axis is the limit of the sum of the products of the masses of 
the elementary particles of which the body is composed by the 
squares of their distances from the axis. 

If we let dM represent any element of mass, and p its distance 
from the axis, then the moment of inertia is 

1= fp 2 dM. 

If we conceive the total mass of the body to be concentrated at 
a single point at a distance, k, from the axis, the moment of iner- 
tia becomes Mk 2 , M being the total mass. 

I 

Hence, k 2 =-jrv- 

k is called the radius of gyration. 

Plane Figures. — The moment of inertia of a plane figure is 
the limit of the sum of the products of the elements of area by 
the squares of their distances from the axis. 

That is, if dA is any element of area, and x is its distance from 
the axis, then the moment of inertia 



■/• 



~-~ K 2 dA=Ak 2 
I 



k 2 — 

K — A 



' Two Parallel Axes. — Fig. 17. Let Z and Z' be two paral- 
lel axes. Then 1 = fp 2 dM and I '= Tp'MM. But d being 

the distance between the axes, and a and b, the co-ordinates of 
Z' referred to O, d 2 =a 2 +b 2 , and 

p 2 =(x - a) 2 -f (y - b) 2 =:x 2 +y 2 +d 2 - 2ax - 2by. 

Therefore I z ' = fpMM -fd 2 f dM -2 a fxdM -2b fydM. (1) 
But jp 2 dM=I , I dM=M > and from the theory of the center 



/cV NOTES ON MECHANICS. 

of gravity, we have j xdM.-.--MX and i ydM MY. 

Therefore, I z '^I z -f M(d 2 -2aX - 2bY) (2) 

in which X and Y are the co-ordinates of the center of gravity of 
the body. 

If Z is a gravity axis, both X and Y become equal to zero, and 
(2) becomes I z ':=I^-|-Md 2 or k a z '— **,,+<*''• (3) 

It is therefore evident that the moment of inertia about a 
gravity axis is less than that about any parallel axis. 

If instead of the element of mass, we use the element of area, 
we obtain for plane figures, I z '=I,-j-Ad' 2 . (4) 

Two Sets of Rectangular Axes with the Same Ori= 
gin. — For two sets of rectangular axes having the same origin 
e may deduce the following relations. 

Fig. 18. Since I x = I y 2 dA, and I Y = | xMA, we have 

ix+^jV-i-r)dA. 

Similarly, lu+I v = f (v 2 +u 2 )dA. 

But since the x and y of any dA have the same hypothenuse as 
the u and v, we have v 2 +u 2 — x 2 -f y 2 . Therefore, I x ; I Y \ u T v . 

Fig. 19. Let X be an axis of symmetry; then, given I x and I Y> 
being anywhere on X, required Iy, U being an axis through O 
and making any angle, a, with X. 

lu— I v 2 dA=- J (y cos a - x sin u)-'dA 

\ u cos'-'a I yMA - 2sina cosa I xydA • sin 2 a I x'dA. 

But since the area is symmetrical about X, in summing up the 
products xydA, for every term x( i y)dA, there is also a term 

x( y)dA to cancel it; therefore f xydA -0. 



NOTES ON MECHANICS. 1$ 

Hence Iy = cos 2 a I x + sin 2 a I Y . 

It may easily be proven that if two distances, a and b, be set off 
from on X and Y respectively, made inversely proportional to 
V I x and i/Iy an d an ellipse be described on a and b as semi-axes ; 
then the moments of inertia of the figure about any axes through 
O are inversely proportional to the squares of the corresponding 
semi-diameters of the ellipse, called therefore the Ellipse of Iner- 
tia. It follows that the moments of inertia about all gravity axes 
of a circle, or of any regular polygon, are equal, since their ellipses 
of inertia must be circles. Even if the plane figure is not sym- 
metrical, an ellipse of inertia can be located at any point, and has 
the properties already mentioned. Its axes are called the princi- 
pal axes for that point. 

The Rectangle. — First, about its base. Fig. 20. Since all 
points of a strip parallel to the base have the same ordinate, we 
may take the area of such a strip as our element of area. 

Then dA=bdz and I B = f z 2 dA = bf Vdz = _Lbh 3 . 

Second, about a gravity axis parallel to the base. Fig. 21. 



= fzMA = b I 



v^:=DCiZ H^rtfOr.: : r: i ; ^ ::: j ! dz = -I bh 3 . 



= "12 



2 

Third, about any other axis in its plane. Use the reduction 
formulas. 

The Triangle. — First, about an axis through the vertex paral- 
lel to the base. Fig. 22. Assume as the element of area a strip 
parallel to the base. Let its length be y, and its breadth dz. Then, 

dA = ydz = c zdz. 
h 



Therefore, I v = J zMA = r ) z ^ Zl 
Second, about a gravity axis parallel to the base. Fig. 23. We 



:-bh 8 . 
4 



20 NOTES ON MECHANICS. 

may obtain the desired result by applying the reduction formula 
for parallel axes to the value just obtained. Since A= bh and 

'^-Ad> = >^bh.fh«=lbh'. 

Third, about the base. I B =I +Ad 2 , with d=-^h. 



3 



Therefore I B =-Lbh 3 -r Ibh.^h 2 = 4bh 3 . 
B 36 2 9 12 



The results in the last two cases could have been obtained di- 
rectly by integration without the use of the reduction formula. 

The Circle. — About any diameter as an axis. Fig. 24. If 
we use polar co-ordinates, the element of area dA=pd<£dp, and 
z =psin<£. 

I g == J zMA= J | (psin<£) 2 /od</>d/>= ( sin'-<pd<£ j pdp 



r 4 /»27T r 4 r2TT \ 

- sin 2 ^>d(/)=- -(1 -cos2<^)d</) 

4J o 4.7 o ^ 



77-r 
~4 



Compound Plane Figures.— Since the moment of inertia 
of any plane figure about any axis is equal to the sum of the mo- 
ments of inertia of its elements, it follows that the moment of 
inertia of any compound plane figure may be found by separating 
it into several simple plane figures whose moments of inertia about 
the given axis may readily be found and then taking the algebraic 
sum of the moments of inertia of these component parts. 

The moment of inertia of the T, shown in Fig. 25, about an 
axis through the base is evidently equal to the moment of inertia 
of the circumscribed rectangle minus the sum of the moments of 
inertia of the two rectangles on either side, i. e., is equal to 

3(bh s -b,V) 



NOTES ON MECHANICS. 21 

To find the moment of inertia of the same figure about the 
gravity axis parallel to the base, we would first find the distance 
from the base to the center of gravity and then apply the reduc- 
tion formula for parallel axes to the result obtained above. 

With this indication of the methods to be followed, the student 
is expected to be able to derive the moments of inertia of other 
compound figures, such as the I, hollow square, annulus, etc., 
about any axis. 

In all of the preceding discussion of moments of inertia of plane 
figures, the axis in every case has been in the plane of the figure. 
Such moments of inertia are called rectangular moments of inertia. 

Polar Moment of Inertia. — If the moment of inertia of a 
plane figure be obtained with reference to an axis perpendicular to 
the plane of the figure, the result is called a polar moment of 
inertia. 

In Fig. 26, let Z be any axis perpendicular to the plane of the 
figure and let X and Y be two rectangular axes in its plane and 
having their origin in Z. Let p be the distance from Z to any ele- 
ment of area, dA. Then I z = I p 2 dA — J (x 2 -f y 2 )dA=I x -j-I Y 

That is, the polar moment of inertia of any plane figure about 
any point in its plane is equal to the sum of the rectangular mo : 
ments of inertia about any two rectangular axes in the plane 
which intersect at that point. 

For the circle about its center, I 7 = ^_. 

z 2 

For a rectangle about its center, I z = -— -bh(b 2 -fh 2 ). 

Thin Plates. Axis in Plate.— Let the plates be homoge- 
neous and of constant, small thickness, t. Let the weight of the 
plate per unit volume be w, and its area, A ; then its mass 

wAt 

is . 

g 



22 NOTES ON MECHANICS. 

Now for a plate whose thickness is very small compared with 
its other dimensions, we may write, 

I - J>dM - ^/>dV =^'J>dA ; 

i. e., =(w^g) X thickness X moment of inertia of the plane 
figure. 

For a rectangular plate about a gravity axis parallel to the base, 

this reduces to I = -t. v bh 3 = -- Mh*. 
8 g 12 12 

The student will be expected to apply this principle to thin plates 

of other shapes. 

Right Prisms of any Altitude about an Axis Perpen- 
dicular to the Base. — As before, the solid is considered homo- 
geneous, its heaviness =w, and its altitude =h. Fig. 27. Con- 
sider an elementary prism whose length is parallel to the axis of 
reference, Z. The altitude of the element will be h, that of the 
whole solid, and its base will be dA, an element of the base, A, of 

*u |.j w whdA 
the solid. Its mass = . 

g 

The moment of inertia of the solid, 

/wh C wh 

p 2 dM = — I />MA = — X the polar moment of inertia of 

the base. 

Applying this to a circular right cylinder, radius of base = r, 

and altitude = h, we have for the moment of inertia about the 

wh 1 1 

axis of the cylinder, I = — .-nx* =^Mr 2 . 
J g g 2 2 

The student is expected to make other applications of this prin- 
ciple. 

Homogeneous Solid Cylinder about a Diameter of 
its Base. — Consider the cylinder to be divided into an infinite 

L.ofC. 



NOTES ON MECHANICS. 23 

number of laminae parallel to the base and of infinitesimal thick- 
ness. Let the distance of any lamina above the base be z, and 
its thickness be dz. Fig. 28. Then the moment of inertia of the 
lamina about its gravity axis parallel to the base of the cylinder is 

w 7rr 4 

— dz. — and the moment of inertia of the lamina about the axis 

g 4 

in the base of the cylinder is this quantity plus the mass of the 
lamina multiplied by the square of its distance above the base. 

The moment of inertia of the cylinder is the sum of the mo- 
ments of inertia of all the laminae, i. e. t 



i x = ri!idz.^+^dz.x^l 

J o *-g 4 g ) 



Homogeneous Right Cone.— First, about an axis through 
the apex and parallel to the base. Fig. 29. Consider the cone 
divided into laminae parallel to the base. Then, if x represents 

r 

the radius of any lamina at a distance, z, from the apex, x=. — z. 

h 



The moment of inertia of any lamina about the assumed axis will 

be wX -I —x 2 + z 2 1 and the moment of inertia of the whole 
g I 4 - | 

cone will be I A = f " xVdz j _L x 2 + z 2 } 

= ™ 2 -!_!.£+ 1 X fzMz=lM(r 2 +4h 2 ). 

Second, about a gravity axis parallel to the base. By use of the 
reduction formula for parallel axes and the result just obtained, we 



have 



^ h 2 

I =_M(r 2 -f-_), 
g 20 ^ 4 



24 NOTES ON MECHANICS. 

Third, about its geometric axis. Since the axis is perpendicu- 
lar to each circular lamina, the moment of inertia of any lamina 

W7rX 2 dz X' 

is equal to its mass X — (radius) 2 = . 

2 g 2 

Now x = — z, hence for the whole cone 
h 

, /~ h W7rr 4 4 , W7rr 2 h 1 3 .. , 

I z = I zMz = . r 2 = Mr 2 . 

J 2gh 4 g 10 10 

Homogeneous Right Pyramid with Rectangular 
Base. — About its geometric axis. Proceeding as in the last case, 

Md 2 

we obtain I = ■- , d being the diagonal of the base. 

Homogeneous Sphere.— About any diameter. Fig. 30. 
Divide into laminae perpendicular to the axis. Noting that 
x 2 = r 2 — z 2 , we have for the whole sphere 

wtt C W7r /~~^ r 

I = - I xMz = _ I (r 4 - 2r 2 z 2 + z f )dz 

Wvffl 2 , , z 5 8 WTrr 5 2 W , 

= — (r 4 z r 2 z 8 -r- -) =-=. = -Mr 2 . 

2gA r 3 5 ; 15 g 5 

For a segment of one or two bases put proper limits for z in 
above instead of -f-r and — r. 



\ 




Fig. 26 



30767 OCT 121903 

iSsr of congress 

003 579 469 4 



